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(C) $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)} \quad K_{c} = 1.844$$	ext{Initial moles at } t = 0: \quad 3.0 \ 	ext{moles of } PCl_{5}$$\

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(C) $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)} \quad K_{c} = 1.844$$	ext{Initial moles at } t = 0: \quad 3.0 \ 	ext{moles of } PCl_{5}$$	ext{Moles at equilibrium } (t = eq): \quad (3-x) \ 	ext{for } PCl_{5}, \ x \ 	ext{for } PCl_{3}, \ x \ 	ext{for } Cl_{2}$$	ext{Equilibrium constant expression: } K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{x^{2}}{3-x} = 1.844$$x^{2} + 1.844x - 5.532 = 0$$	ext{Using quadratic formula: } x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a} = \frac{-1.844 + \sqrt{(1.844)^{2} + 4(5.532)}}{2}$$x = \frac{-1.844 + \sqrt{3.400 + 22.128}}{2} = \frac{-1.844 + 5.052}{2} \approx 1.604$$	ext{Moles of } PCl_{5} 	ext{ at equilibrium} = 3 - 1.604 = 1.396 \ 	ext{moles}$$1.396 \ 	ext{moles} = 1396 	imes 10^{-3} \ 	ext{moles} \approx 1400 	imes 10^{-3} \ 	ext{moles}$

$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \quad K_{c} = 1.844$
$3.0 \ \text{moles}$ of $PCl_{5}$ is introduced in a $1 \ \text{L}$ closed reaction vessel at $380 \ \text{K}$. The number of moles of $PCl_{5}$ at equilibrium is $..... \times 10^{-3}$. (Round off to the Nearest Integer)

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$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \quad K_{c} = 1.844$$3.0 \ \text{moles}$ of $PCl_{5}$ is introduced in a $1 \ \text{L}$ closed reaction vessel at $380 \ \text{K}$. The number of moles of $PCl_{5}$ at equilibrium is $..... \times 10^{-3}$. (Round off to the Nearest Integer)